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Let $F(x)=f(x)+f\left(\frac{1}{x}\right)$, where $f(x)=\int_1^x \frac{\log t}{1+t} d t$. Then $F(e)$ equals
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$\frac{1}{2}$
$\frac{1}{2}$
$f(x)=\int_1^x \frac{\log t}{1+t} d t$
$F(e)=f(e)+f\left(\frac{1}{e}\right)$
$F(e)=\int_1^e \frac{\log t}{1+t} d t+\int_1^{1 / e} \frac{\log t}{1+t} d t$
$=\int_1^e \frac{\log t}{1+t}+\int_1^e \frac{\log t}{t(1+t)} d t$
$=\int_1^e \frac{\log t}{t} d t=\frac{1}{2}$.
$F(e)=f(e)+f\left(\frac{1}{e}\right)$
$F(e)=\int_1^e \frac{\log t}{1+t} d t+\int_1^{1 / e} \frac{\log t}{1+t} d t$
$=\int_1^e \frac{\log t}{1+t}+\int_1^e \frac{\log t}{t(1+t)} d t$
$=\int_1^e \frac{\log t}{t} d t=\frac{1}{2}$.
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